Ta có: \(VT=\sqrt{\left(2n+1\right)^2}+\sqrt{4n^2}=\sqrt{\left(2n+1\right)^2}+\sqrt{\left(2n\right)^2}\)

\(=\left|2n+1\right|+\left|2n\right|\)

Vì \(n\inℕ\)\(\Rightarrow2n+1>0\); \(2n\ge0\)

\(\Rightarrow\left|2n+1\right|=2n+1\)và \(\left|2n\right|=2n\)

\(\Rightarrow VT=2n+1+2n=4n+1\)

Ta có: \(VP=\left(2n+1\right)^2-4n^2=\left(2n+1\right)^2-\left(2n\right)^2\)

\(=\left(2n+1-2n\right)\left(2n+1+2n\right)=4n+1\)

\(\Rightarrow VT=VP\)\(\Rightarrowđpcm\)

a, 2020 lớn hơn

a)\(\left(\sqrt{2019.2021}\right)^2=2019.2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1< 2020^2\)

=> \(\sqrt{2019.2021}< 2020\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>5+2\sqrt{4}=5+2.2=9\)

=> \(\sqrt{2}+\sqrt{3}>3\)

c) \(9+4\sqrt{5}=4+4\sqrt{5}+5=\left(2+\sqrt{5}\right)^2>\left(2+\sqrt{4}\right)^2=\left(2+2\right)^2=16\)

=> \(9+4\sqrt{5}>16\)

d) \(\sqrt{11}-\sqrt{3}>\sqrt{9}-\sqrt{1}=3-1=2\)

=> \(\sqrt{11}-\sqrt{3}>2\)

Đề sai r bạn phải là \(2020\sqrt{2019}\)

ko bt tự làm đi!!
 

ko bt nha

câu này dễ

trả lời : ko bt :)

????? nha ban

Xét: \(A^3=x^3+3A\sqrt[3]{\frac{4}{4}}\Leftrightarrow A^3=x^3-3x+3A\Leftrightarrow A^3-3A-x^3+3x=0\)

\(\Leftrightarrow\left(A-x\right)\left(A^2+Ax+x^2\right)-3\left(A-x\right)=0\)\(\Leftrightarrow\left(A-x\right)\left(A^2+Ax+x^2-3\right)=0\)

\(\cdot A-x=0\Leftrightarrow A=x=\sqrt[3]{1995}\)

\(\cdot A^2+Ax+x^2-3=0\) có \(\Delta=3\left(4-x^2\right)< 0\)vì \(x=\sqrt[3]{1995}\)

Do đó phương trình cuối vô nghiệm. Vậy \(A=\sqrt[3]{1995}\)

a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{3-2\sqrt{3}+1}-\sqrt{3}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)

b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{9+6\sqrt{2}+2}-3+\sqrt{2}\)

\(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)

c) \(\sqrt{25x^2}-2x=-5x-2x=-7x\)(vì x < 0)

d) \(x-5+\sqrt{25-10x+x^2}=x-5+\sqrt{\left(5-x\right)^2}=x-5+x-5=2x-10\) (vì x > 5)